After you substitute for $y$, you will see an $x^4$ term. Try expanding carefully and then look for a way to rewrite the equation using $t = x^2$ so it becomes a simpler quadratic.

When you solve the quadratic in $t$, remember that $t = x^2$. Which of the solutions for $t$ make sense for a square? How many values of $x$ come from each acceptable $t$?

Once you know all real $x$ that work, plug them back into $y = x^2 - 2$. How many different ordered pairs $(x, y)$ does that give you?

In Desmos, enter the equation x^2 + y^2 = 10 to graph the circle centered at the origin with radius $\sqrt{10}$.

On a new line, enter y = x^2 - 2 to graph the parabola that opens upward and is shifted down by 2 units.

Click on the points where the circle and the parabola intersect. Desmos will highlight each intersection point and show its coordinates.

Count how many distinct intersection points appear. Each intersection corresponds to a real ordered pair $(x, y)$ that satisfies the system; this count is the answer.

Use the second equation, $y = x^2 - 2$, and substitute it into the first equation $x^2 + y^2 = 10$.

That gives:

Now the equation involves only $x$, so we can solve for $x$.

Expand and simplify the equation:

So

Let $t = x^2$. Then the equation becomes a quadratic:

Now solve this quadratic for $t$.

Use the quadratic formula on $t^2 - 3t - 6 = 0$:

So the two possible values for $t$ (which equals $x^2$) are

Since $x^2 \ge 0$, only nonnegative $t$ values are allowed.

Note that $\sqrt{33} > 5$, so $3 - \sqrt{33}$ is negative and therefore $t_2$ is negative, which cannot equal $x^2$.

Thus there is exactly one valid value of $x^2$, namely $x^2 = \dfrac{3 + \sqrt{33}}{2}$.

From $x^2 = \dfrac{3 + \sqrt{33}}{2}$, there are two real $x$ values:

For each $x$, compute $y$ using $y = x^2 - 2$:

So the two solution pairs are

There are two distinct real ordered pairs $(x, y)$ that satisfy the system, so the correct choice is Exactly two.